Total no of terms are (2n+1) which are odd (right?), so middle term will be (2n+1 +1) /2 th term i.e. (n+1)th term

Now middle term i.e. (n+1) th term of (1+x)^{2n} = ^{2n}C_{n} (x)^{n, }= !2n /(!n *!n) = 1.2.3.4.......(2n-1) (2n)x^{n}/ (!n * !n)

(remember 1.2.3.....(2n) are total '2n' terms out of which 'n' will be even (i.e. 2.4.6.....upto (2n) ) and 'n' are odd (i.e. 1.3.5.7..upto (2n-1) ).

1.2.3.4.......(2n-1) (2n)/ (!n * !n) = 1.3.5.7...(2n-1) * (2.4.6.....2n)x^{n} /!n *!n

Taking out '2' each from these 'n' 'even' termswe get 1.3.5...(2n-1) (2^{n})(1.2.3.4.5.....n) x^{n}/!n *!n,

or ,1.3.5...(2n-1) (2^{n}) !n/(!n*!n) which gives 1.3.5...(2n-1) (2^{n})x^{n}/ !n